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An electric potential of $9 \times 10^5\; V$ is caused at point P by two particles of charges $50 \mu C$ and $−20 \mu C$, as shown in the given figure.The point P is situated from the negative charge at a distance of

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(D) 10.49 m
Hence D is the correct answer.
answered Jun 10, 2014 by meena.p
 

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